Wythoff construction

Remarks

All points of $W v$ lie on the sphere of radius $\norm{v}$, so each of them is a vertex of $P(v)$, and $W$ acts transitively on the vertices of $P(v)$. The polytope depends, up to similarity, mainly on the set of mirrors $H_i$ that do not contain $v$; in a Coxeter diagram one marks the corresponding nodes with rings. If $v$ does not lie on the mirror $H_i$, then $v$ and $s_i(v)$ are joined by an edge of length twice the distance from $v$ to $H_i$. When $v$ has the same distance to every mirror that does not contain it, all edges of $P(v)$ have the same length, and $P(v)$ is a uniform polytope. Let $W$ be the symmetry group of a regular polytope. If $v$ lies on all mirrors except the one at an end of the Coxeter diagram, then $P(v)$ is that regular polytope or its dual, up to similarity.

Example: let $W$ be the symmetry group of the square, of order $8$, generated by the reflections in the lines $y = 0$ and $y = x$, with chamber $C = \set{(x, y) : 0 \le y \le x}$. For $v = (1, 0)$, on the mirror $y = 0$, the orbit is $\set{(\pm1, 0), (0, \pm1)}$, and $P(v)$ is a square. For $v = (1, 1)$, on the mirror $y = x$, the orbit is $\set{(\pm1, \pm1)}$, again a square. For $v = (1, \tan(\pi/8))$, which is at equal distance from both mirrors, the orbit has $8$ points spaced by angles $\pi/4$, and $P(v)$ is a regular octagon.

Non-example: the choice $v = 0$ is excluded; its orbit is the single point $0$, and the construction gives no polytope of dimension $n$.